5% Search

Q:

A rectangular courtyard 3.78 meters long 5.25 meters wide is to be paved exactly with square tiles, all of the same size. what is the largest size of the tile which could be used for the purpose?

A) 14 cms	B) 21 cms
C) 42 cms	D) None of these

Answer & Explanation Answer: B) 21 cms

Explanation:

3.78 meters =378 cm = 2 × 3 × 3 × 3 × 7

5.25 meters=525 cm = 5 × 5 × 3 × 7

Hence common factors are 3 and 7

Hence LCM = 3 × 7 = 21

Hence largest size of square tiles that can be paved exactly with square tiles is 21 cm.

Report Error

View Answer Report Error Discuss

Filed Under: HCF and LCM

Q:

The H.C.F and L.C.M of two numbers are 11 and 385 respectively. If one number lies between 75 and 125 , then that number is

A) 77	B) 88
C) 99	D) 110

Answer & Explanation Answer: A) 77

Explanation:

Product of numbers = 11 x 385 = 4235

Let the numbers be 11a and 11b . Then , 11a x 11b = 4235 => ab = 35

Now, co-primes with product 35 are (1,35) and (5,7)

So, the numbers are ( 11 x 1, 11 x 35) and (11 x 5, 11 x 7)

Since one number lies 75 and 125, the suitable pair is (55,77)

Hence , required number = 77

Report Error

View Answer Report Error Discuss

Filed Under: HCF and LCM

Q:

If the sum of two numbers is 55 and the H.C.F and L.C.M of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to

A) 55/601	B) 601/55
C) 11/120	D) 120/11

Answer & Explanation Answer: C) 11/120

Explanation:

Let the numbers be a and b . Then, a+b =55 and ab = 5 x 120 = 600.

Therefore, Required sum = $\frac{1}{a} + \frac{1}{b} = \frac{a + b}{a b} = \frac{55}{600} = \frac{11}{120}$

Report Error

View Answer Report Error Discuss

Filed Under: HCF and LCM

Q:

The L.C.M of two numbers is 495 and their H.C.F is 5. If the sum of the numbers is 100, then their difference is

A) 10	B) 46
C) 70	D) 90

Answer & Explanation Answer: A) 10

Explanation:

Let the numbers be x and (100-x).

Then, $x (100 - x) = 5 * 495$

=> $x^{2} - 100 x + 2475 = 0$

=> (x-55) (x-45) = 0

=> x = 55 or x = 45

The numbers are 45 and 55

Required difference = (55-45) = 10

Report Error

View Answer Report Error Discuss

Filed Under: HCF and LCM

Q:

The H.C.F of two numbers is 11 and their L.C.M is 7700. If one of the numbers is 275 , then the other is

A) 279	B) 283
C) 308	D) 318

Answer & Explanation Answer: C) 308

Explanation:

Other number = $(\frac{11 \times 7700}{275})$ = 308

Report Error

View Answer Report Error Discuss

Filed Under: HCF and LCM

Q:

The sum of two numbers is 528 and their H.C.F is 33. The number of pairs of numbers satisfying the above condition is

A) 4	B) 6
C) 8	D) 12

Answer & Explanation Answer: A) 4

Explanation:

Let the required numbers be 33a and 33b.

Then 33a +33b= 528 => a+b = 16.

Now, co-primes with sum 16 are (1,15) , (3,13) , (5,11) and (7,9).

Therefore, Required numbers are ( 33 x 1, 33 x 15), (33 x 3, 33 x 13), (33 x 5, 33 x 11), (33 x 7, 33 x 9)

The number of such pairs is 4

Report Error

View Answer Report Error Discuss

Filed Under: HCF and LCM

Q:

The L.C.M of 22, 54, 108, 135 and 198 is

A) 330	B) 1980
C) 5940	D) 11880

Answer & Explanation Answer: C) 5940

Explanation:

22 = 2 x 11

54 = $2 \times 3^{3}$

108 = $2^{2} \times 3^{3}$

135 = $3^{3} \times 5$

198 = $2 \times 3^{2} \times 11$

$∴ L . C . M = 2^{2} \times 3^{3} \times 5 \times 11 = 5940$

Report Error

View Answer Report Error Discuss

Filed Under: HCF and LCM

Q:

H.C.F of 4 x 27 x 3125, 8 x 9 x 25 x 7 and 16 x 81 x 5 x 11 x 49 is

A) 180	B) 360
C) 540	D) 1260

Answer & Explanation Answer: A) 180

Explanation:

$4 \times 27 \times 3125 = 2^{2} \times 3^{3} \times 5^{5}$

$8 \times 9 \times 25 \times 7 = 2^{3} \times 3^{2} \times 5^{2} \times 7$

$16 \times 81 \times 5 \times 11 \times 49 = 2^{4} \times 3^{4} \times 5 \times 7^{2} \times 11$

Therefore, H.C.F = $2^{2} \times 3^{2} \times 5$ = 180

$\inline \fn_jvn \therefore$ $\inline \fn_jvn H.C.F =2^{2}\times 3^{2}\times 5=180$

Report Error

View Answer Report Error Discuss

Filed Under: HCF and LCM

Searching for "5%"

Quick Links